Which of the following is a galvanic cell?
2 answers:
C. Aluminum (Al) oxidized, zinc (Zn) reduced
<h3>Further explanation</h3>Given
Metals that undergo oxidation and reduction
Required
A galvanic cell
Solution
The condition for voltaic cells is that they can react spontaneously, indicated by a positive cell potential.
or:
E ° cell = E ° reduction-E ° oxidation
For the reaction to occur spontaneously (so that it E cell is positive), the E° anode must be less than the E°cathode
If we look at the voltaic series:
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The standard potential value(E°) from left to right in the voltaic series will be greater, so that the metal undergoing an oxidation reaction (acting as an anode) must be located to the left of the reduced metal (as a cathode)
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From the available answer choices, oxidized Al (anode) and reduced Zn (cathode) are voltaic/galvanic cells.
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Answer: -
15.55 M
35.325 molal
Explanation: -
Let the volume of the solution be 1000 mL.
Density of nitric acid = 1.42 g/ mL
Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid
= 1000 mL x 1.42 g/ mL
= 1420 g.
Percentage of HNO₃ = 69%
Amount of HNO₃ =
= 979.8 g
Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol
Number of moles of HNO₃ =
= 15.55 mol
Molarity is defined as number of moles per 1000 mL
We had taken 1000 mL as volume and found it to contain 15.55 moles.
Molarity of HNO₃ = 15.55 M
Mass of water = Total mass of nitric acid solution - mass of nitric acid
= 1420 - 979.8
= 440.2 g
So we see that 440.2 g of water contains 15.55 moles of HNO₃
Molality is defined as number of moles of HNO₃ present per 1000 g of water.
Molality of HNO₃ =
= 35.325 molal
Answer: 2.4 ml
Solution :
Molar mass of = 17 g/mole
Given,: 28% w/w of solution means 28 g of ammonia in 100 g of solution.
Mass of solution = 100 g
Now we have to calculate the volume of solution.
Molarity : It is defined as the number of moles of solute present in one liter of solution.
where,
n = moles of solute
= volume of solution in liter = 0.11 L
Now put all the given values in the formula of molarity, we get
Using molarity equation: